H. A. TANAKA LECTURE 6: MORE PROPERTIES OF THE DIRAC EQUATION TRANSFORMING THE DIRAC EQUATION:
µ µ i ⇤µ⇥ mc ⇥ =0 i ⇤ ⇥ mc ⇥ =0 ⇥ µ µ i ⇤ (S⇥) mc (S⇥)=0 µ ⇥ µ ⇤x i µ ⇤⇥ (S⇥) mc (S⇥)=0 ⇤x S is constant in space time, so we can move it to the left of the derivatives ⇥ µ ⇤x i S µ ⇤⇥ ⇥ mc (S⇥)=0 ⇤x Now slap S-1 from both sides @x⌫ i ⇤ ⇥ mc ⇥ =0 1 µ S i~ S @⌫ mc S =0 ! @xµ0 Since these equations must be the same, S must satisfy ⇥ ⇥ 1 µ ⇥x = S S µ ⇥x ⇥ EXAMPLE: THE PARITY OPERATOR
• For the parity operator, we want to invert the spatial coordinates while keeping the time coordinate unchanged:
x x1 10 0 0 0 =1 = 1 0 10 0 x0 x1 P = ⇥ 00 10 x x2 3 ⇧ 00 0 1 ⌃ = 1 = 1 ⇧ ⌃ x2 x3 ⇤ ⌅
• We then have Recalling We find that γ0 satisfies our needs ⇥ 0 1 0 ⇥ 1 µ ⇥x = S S = S S ⇥xµ 0 = 0 0 0 = 0 1 1 1 ⇥ = S S 0 10 i 0 i 0 0 0 i i 2 1 2 = = = = = S S 0 1 3 1 3 ⇥ = S S 0 2 ( ) =1 0 µ, ⇥ =2gµ⇥ 0 i = i 0 SP = { } ⇥ LORENTZ COVARIANT QUANTITIES
• Recall that for four vectors aµ, bµ • aµ, bµ transform as Lorentz vectors (obviously) µ • a bµ is a scalar (does not change under Lorentz transformations • aµbν is a tensor (each has a Lorentz transformation) • From the previous discussion, we know: • Dirac spinors have four components, but don’t transform as Lorentz vectors • How do combinations of Dirac spinors change under Lorentz Transformations? HOW DO WE CONSTRUCT A SCALAR? 0 ¯ 0 • We can use γ : define: ⇥ = ⇥† • Consider a Lorentz transformation with S acting on the spinor 0 0 • We can also show generally that S† S = 0 0 • This gives us ⇥⇥¯ ⇥†S† S⇥ = ⇥† ⇥ = ⇥⇥¯ • so this is a Lorentz invariant • We can construct the parity operator to check how ¯ transforms under the parity operation. 0 • Recall SP = γ • We can investigate how ¯ transforms under parity 0 0 0 0 0 ⇥⇥¯ (⇥†S† )(S ⇥)=⇥† ⇥ = ⇥† ⇥ = ⇥⇥¯ P P ¯ doesn’t change sign under parity it is a Lorentz scalar THE γ 5 OPERATOR
5 • Define the operator γ as: 01 5 0 1 2 3 10 = i ⇥ • It anticommutes with all the other γ matrices: µ, 5 =0
⇥ µ • use the canonical anti-commutation relations to move γ to the other side µ • γ will anti-commute with for µ≠ν µ • γ will commute when µ=ν ¯ 5 • We can then consider the quantity ⇥ ⇥ • Can show that this is invariant under Loretnz transformation. • What about under parity? 5 0 5 0 5 0 5 5 ⇥ ¯ ⇥ (⇥†S† ) (S ⇥)= (⇥†S† ) S ⇥ = ⇥† ⇥ = ⇥ ¯ ⇥ ⇥ P P P P ⇥ ¯ 5 ⇥ is a “pseudoscalar” OTHER COMBINATIONS
µ • We can use γ to make vectors and tensor quantities: ⇤⇤¯ scalar 1 component ⇤ ¯ 5⇤ pseudoscalar 1 component ⇤ ¯ µ⇤ vector 4 components ⇤ ¯ µ 5⇤ pseudovector 4 components i ⇤⇥¯ µ⇥ ⇤ antisymmetric tensor 6 components ⇥µ⇥ = ( µ ⇥ ⇥ µ) 2 • You can tell the transformation properties by looking at the Lorentz indices 5 • γ introduces a sign (adds a “pseudo”) * • Every combination of ψ iψj is a linear combination of the above. • We will see that the above are the basis for creating interactions with Dirac particles. Interactions will be classified as “vector” or “pseudovector”, etc. Angular Momentum and the Dirac Equation:
• Conservation: • In quantum mechanics, what is the condition for a quantity to be conserved? [H, Q]=0 • Free particle Hamiltonian in non-relativistic quantum mechanics: p2 p2 1 H = [H, p]=[ ,p]= [p2,p]=0 2m 2m 2m • thus we conclude that the momentum p is conserved • If we introduce a potential: p2 1 [H, p]=[ + V (x),p]= [p2 + V (x),p] =0 2m 2m
• thus, momentum is not conserved Hamiltonian for the Dirac Particle:
• Starting with the Dirac equation, ( µp mc)⇥ =0 µ • determine the Hamiltonian by solving for the energy
• Hints: ( 0)2 =0
• Answer: H = c 0 ( p + mc) · Is orbital angular momentum conserved?
• We want to evaluate [H, L ]
• Recall: j k L = x p Li = ijkx p 0 0 a b H = c ( p + mc) H = c ⇥ab p + mc · 0 a b j k [H, Li]=[c ⇥ab p + mc , ⇤ijkx p ] • which parts do not commute?
0 a b [c ⇥ab p , ⇤ijkxjpk] [mc, ijkxjpk] • 0 b j k c ⇥ab⇤ijk[p ,x p ] [A, BC]=[A, B]C + B[A, C]
0 a bj k 0 j k 0 c⇥ab⇤ijk ( i ⇥ p ) i c ⇥ijk p i c (⇥ p⇥) Orbital angular momentum is not conserved Spin
⇥ 0 • Consider the operator:S⇥ = ⇥ = acting on Dirac spinors 2 2 0 ⇥ • Note that it satisfies all the properties of an angular momentum operator. • Let’s consider the commutator of this with the hamiltonian H = c 0 ( p + mc) ~c · [ 0~ p~ + 0mc, ⌃~ ] 2 · • once again, consider in component/index notation ~c [H, Si]= [ 0 apb + 0mc, ⌃i] 2 ab • which part doesn’t commute? c ⇥ pb[ 0 a, i] [AB, C]=[A, C]B + A[B,C] 2 ab
a i c b 0 a i 0 0 ⇥abp [ , ] [ , ] 2 a 0 0 i
j 0[ , ] 0 j a i aij [ a, ]0 aij j 0 i The “total” spin operator:
• Define the operator: 0 2 S = = S = S S 2 0 · ⇥
• Calculate its eigenvalue for an arbitrary Dirac spinor:
• If the eigenvalue of the operator gives s(s+1), where s is the spin, what is the spin of a Dirac particle? NEXT TIME • Please turn in problem set 1 by 1600 on Thursday (6 Oct) • Please read Chapter 5